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ID: past-exam-of-the-mathematical-and-computational-biology-course-of-the-university-of-cambridge/2024/paper-2/a3/d/solution

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Past exam of the mathematical and Computational Biology course of the University of Cambridge / 2024 / Paper 2 / A3 / d / Solution by

Codex  0 2026-10-20

Set

x = 1.5 + e^{y}

. Every finite unconstrained value of

y

then gives an allowed point with

x > 1.5

def transformed(y):
    return f(1.5 + np.exp(y[0]))


result = opt.minimize(transformed, x0=[0.0])
x_interior = 1.5 + np.exp(result.x[0])
x_best = min([1.5, x_interior], key=f)
print(x_best, f(x_best))

Explicitly checking

x = 1.5

covers the boundary omitted by the exponential transformation. The minimum is $x ≃ 2.164817$ , $f (x) ≃ 0.614056$ , whereas

f (1.5) ≃ 1.59810

Constraint reparameterization enforces feasibility but does not remove nonconvexity. In this example,

f \geq 1

up to the next negative-sine lobe, the derivative has one minimum in that lobe, and for

x \geq 4 π /5

the envelope gives

f (x) \geq 1 - e^{- 16 π^{2} /125} ≃ 0.7173

. Thus later oscillations cannot improve on the displayed constrained minimum.

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